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Linear Spaces (Vectors)

  1. Jan 29, 2009 #1
    1. The problem statement, all variables and given/known data
    From the course of Linear Algebra and Analytic Geometry

    I need to find the dimension and two different bases of subspace R3 generated by vectors (1,2,3), (4,5,6), (7,8,9).

    2. Relevant equations
    None.

    3. The attempt at a solution
    I tried

    (a,b,c)=α1(1,2,3)+α2(4,5,6)+α3(7,8,9)

    which became

    a = α1 + 4α2 + 7α3
    b = 2α1 + 5α2 + 8α3
    c = 3α1 + 6α2 + 9α3

    which (by Gaussian elimination) became an undetermined system with free variable α3.

    4. The solution given by teacher
    dim=2, example of bases={(1,0,-1),(0,1,2)} or {(2,1,0),(-1,0,1)}

    I don't want the solution, I just want to understand the mechanics on how to find the bases and the generated subspace. If someone could explain it to me, thank you.
     
  2. jcsd
  3. Jan 29, 2009 #2

    Mark44

    Staff: Mentor

    The system of equations you show would be used to find the span of your three vectors. Instead of equations that start with a= , b=, and c=, put 0 in for all three of those variables. You should end up with a row of zeroes and two nonzero rows.

    What did you end up with when you row-reduced your matrix?
     
  4. Jan 29, 2009 #3
    So, I should have used the homogeneous system A . x = 0 ? Being A a matrix. Hmm...
    I ended up with (after Gaussian elimination):

    Code (Text):
    [ 1  4  7 | a      ]
    [ 0 -3 -6 | b-2a   ]
    [ 0  0  0 | c-a-2b ]
     
  5. Jan 29, 2009 #4

    Mark44

    Staff: Mentor

    If I can backpedal a bit, your work is fine. For the system represented by your augmented matrix to be consistent, it must be that c - a - 2b = 0.

    or
    Code (Text):

    a = -2b + c
    b =    b
    c =          c
     
    I added the 2nd and 3rd equations above so that I can get some vectors out of the equation c - a - 2b = 0. The equations I added are obviously true for all values of b and c, respectively.

    Any vector [a b c]^T is a linear combination of [-2 1 0]^T and [1 0 1]^T. These come from setting b = 1, c = 0 and then b = 0, c = 1.

    Different pairs of choices for b and c will give you different pairs of vectors for your basis.

    Hope that helps.
     
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